Determine if the sets of vectors are a basis for the indicated space

Question

I am confused on how to determine if the sets of vectors are a basis of the space:

$a) \{(5,2,-1), (1,0,1), (3,-1,0)\}$ in $\mathbb R^3$ , and

$b) \{1,0,,-2,5), (4,4,-3,2), (0,1,0,-3), (1,3,3,-10)\}$ in $\mathbb R^4$.

Attempt:

$a)$ I put each of the vectors into an equation as follows:

$$5a+b+3c=0 \\ 2a-c=0 \\ -a+b=0$$

and found out that the equations above will only be true if $a=b=c=0,$ and therefore they are linearly independent. I also found out that the matrix is invertible through calculating the determinant (which equals $12$), and therefore these sets of vectors are a basis of $\mathbb R^3$.

$b)$ I put each of the vectors into a matrix, and after RREF I found out that there are solutions to each of the vectors, but do not know how to carry on.

Am I doing something wrong in my work or am I giving a false conclusion?

Answer 1

For the first set of vectors the determinant is 6 (not 0) which indicates that the matrix is inversible, thus the vectors are linearly independent, and these 3 vectors FORM a base of $\mathbb R^3$.

For the set b, you can simply put the vectors in a 4x4 matrix and calculate the determinant, you will get 0 indicating that the vectors aren’t linearly independent and thus they can’t form a base of $\mathbb R^4$.

You can also proceed in the same manner of a, by setting the vectors in a system of 4 unknowns $(a,b,c,d)$ and solve it for $(0,0,0,0)$, you will definitely get an answer other than vector 0 indicating the dependence between these vectors.

Hope I helped.

Answer 2

In both case the most efficent method is to arrange the vectors as rows in a matrix (note that it is not defined augmented) and obtain the RREF by elementary row operations.

From the number of pivots in the RREF we can find if the vectors are linearly independent.

Answer 3

For your first question, upon reaching the conclusion that $a=b=c=0$, you can conclude that it forms a basis. You have calculated the determinant wrongly, it should not be zero.

> det(matrix(c(5,2,-1,1,0,1,3,-1,0),nrow=3, byrow=TRUE))
[1] 12

You should repeat the same procedure for the second question, if you can find a non-trivial solution, then it is linearly dependent and it is not a basis. Alternatively, just compute the determinant and check if it is zero.(The second answer is not a basis).